Definition 10.1 Let \(z_0\) be an isolated singularity of a holomorphic function \(f\colon U\to\C.\) The residue \(\Res_{z_0}(f)\) is the residue of the Laurent series Equation 9.8. In other words, \[f(z)=\sum_{n=-\iy}^\iy a_n(z-z_0)^n\implies \Res_{z_0}(f)=a_{-1}.\]
Proposition 10.1 Let \(z_0\) be an isolated singularity of both the holomorphic functions \(f,g\colon U\to\C.\)
Proof.
and b.are left as an Exercise.
If \(f=\sum_{n=-m}^\iy a_n(z-z_0)^n,\) then \(h(z)=\sum_{n=-m}^\iy a_n(z-z_0)^{n+m}\) is a power series which we can differentiate termwise by Theorem 5.2 to get \[h^{(m-1)}(z)=\sum_{n=-m}^\iy (n+m)(n+m-1)\cdots(n+2) a_n(z-z_0)^{n+1}.\]
The limit \(z\to z_0\) of this power series is its constant term, \((m-1)!a_{-1}.\)
Example 10.1
\(\Res_0(z^n)=\begin{cases} 0 & \text{if $n\neq-1$},\\ 1 & \text{if $n=-1$}. \end{cases}\)
Example 10.2
The best way to compute a residue is often to work out the Laurent expansion using the calculus of formal power series from Chapter 5.
Consider \(f(z)=\frac{\sin(z^2)}{z^7}.\) As \(\sin(z)=\sum_{n=0}^\iy\frac{(-1)^n}{(2n+1)!}z^{2n+1},\) we have \[f(z)=\sum_{n=0}^\iy\frac{(-1)^n}{(2n+1)!}z^{4n+2-7},\] hence \(\Res_0f(z)=-1/6\) is the coefficient of the term with \(n=1.\)
Let \(g(z)=\frac{\cos(z)-1}{(\exp(z)-1)^3}.\) It suffices to work with finite portions of the involved power series. For this, introduce the notation \(O(z^m)\) to mean ‘terms involving \(z^n\) of order \(m\leqslant n\)’. We will use this in an informal way, which can be made precise in algebra by saying that \(O(z^n)\) is the ideal in the ring of formal power series generated by \(z^n.\) Notice the properties \[\begin{align} O(z^m)O(z^n)&=O(z^{m+n}),\\ O(z^m)^n&=O(z^{mn}),\\ O(z^m)+O(z^n)&=O\bigl(z^{\min(m,n)}\bigr),\\ 1/O(z^n)&=O(z^{-n}). \end{align}\] Now compute \[\begin{align*} \exp(z)-1&=z+z^2/2+O(z^3),\\ \implies(\exp(z)-1)^3&=\frac{z^6}{8}+\frac{3 z^5}{4}+\frac{3 z^4}{2}+z^3+O(z^5)\\ &=z^3+\frac{3z^4}{2}+O(z^5). \end{align*}\] To find the expansion of the inverse \(1/(\exp(z)-1)^3,\) we apply the proof strategy of Proposition 5.1(b) to \(z^{-3}(\exp(z)-1)^3=1+\frac{3z}{2}+O(z^2).\) In other words, we compare coefficients to solve \[\left(1+\frac{3z}{2}+\ldots\right)(a_0+a_1z+a_2z^2+\ldots)=1\]
recursively for \(a_0, a_1, a_2.\) This gives \[1/(\exp(z)-1)^3=z^{-3}\left(1-\frac{3z}{2}+\frac{9z^2}{4}+O(z^3)\right)=z^{-3}-\frac{3z^{-2}}{2}+\frac{9z^{-1}}{4}+O(z^0).\]
Finally, multiply this by \(\cos(z)-1=-\frac{z^2}{2}+\frac{z^4}{24}+O(z^5)\) to get \[g(z)=-\frac{1}{2 z}+\frac{3}{4}-\frac{13 z}{12}-\frac{z^2}{16}+O(z^2)\]
from which we read off \(\Res_0g(z)=-1/2.\)
Definition 10.2 Let \(\ga\colon[a,b]\to\C\) be a closed piecewise C1 curve and let \(z_0\in\C\setminus\ga([a,b]).\) The winding number of \(\ga\) around \(z_0\) is\[W_{z_0}(\ga)=\frac{1}{2\pi i}\int_\ga\frac{dz}{z-z_0}.\]
Example 10.3
Let \(\ga\colon[0,2\pi]\to\C,\) \(\ga(t)=z_0+re^{it},\) be the boundary curve of the disk \(\ol D_r(z_0).\) Then Equation 6.6 implies \(W_{z_0}(\ga)=1.\) More generally, for every \(w_0\in D_r(z_0)\) the Cauchy integral formula Equation 8.1 gives\[W_{w_0}(\ga)=1.\]
Pick a subdivision \(a=t_0<t_1<\cdots<t_n=b\) such that \(\ga|_{[t_{k-1},t_k]}\) is continuously differentiable. Using branches of the logarithm, we can write \(\ga(t)=z_0+r(t)e^{i\th(t)}\) for \(r(t)=|\ga(t)-z_0|\) and a continuous function \(\th\colon[a,b]\to\R\) with \(\th|_{[t_{k-1},t_k]}\) continuously differentiable. With this notation, the following holds.
Proposition 10.2
Proof.
Over \([t_{k-1},t_k]\) we have \(\ga'(t)=r'(t)e^{i\th(t)}+r(t)\th'(t)e^{i\th(t)}.\) Using\[\frac{\ga'(t)}{\ga(t)-z_0}=\frac{r'(t)}{r(t)}+i\th'(t) \tag{10.1}\] we compute \[2\pi i W_{z_0}(\ga)=\sum_{k=1}^n\int_{t_{k-1}}^{t_k}\frac{\ga'(t)}{\ga(t)}dt\] by Equation 10.1 we have \[\sum_{k=1}^n\int_{t_{k-1}}^{t_k}\frac{\ga'(t)}{\ga(t)}dt = \sum_{k=1}^n\left(\int_{t_{k-1}}^{t_k}\frac{d}{dt}\log r(t)dt + i\int_{t_{k-1}}^{t_k}\th'(t)dt\right)\] using FTC, the right hand side becomes \[\sum_{k=1}^n \Bigl(\log r(t_k)- \log r(t_{k-1})+i\th(t_k)-i\th(t_{k-1})\Bigr)\] observing that these are telescoping sums, we conclude that \[2\pi i W_{z_0}(\ga)= i\bigl(\th(t_n)-\th(t_0)\bigr)=i\bigl(\th(b)-\th(a)\bigr)\]and the claimed formula follows. From \(\ga(a)=\ga(b)\) we have \(e^{i\th(a)}=e^{i\th(b)},\) so Proposition 1.5 implies \(\th(b)-\th(a)=2\pi k\) for some \(k\in\Z.\)
By Theorem 7.2 applied to \(U=\C\setminus\{z_0\}\) and \(f(z)=1/(z-z_0),\) the implication ‘\(\Longleftarrow\)’ holds.
Conversely, assume \(W_{z_0}(\ga_0)=W_{z_0}(\ga_1).\) Write \(\ga_k(t)=z_0+r_k(t)e^{i\th_k(t)}\) for \(k=0,1.\) As \(\ga_k(a)=\ga_k(b),\) we have \(r_k(a)=r_k(b)\) and \(\th_k(a)=\th_k(b)-2\pi i n_k\) for some \(n_k\in\Z.\) By assumption, \(\frac{\th_0(b)-\th_0(a)}{2\pi}=\frac{\th_1(b)-\th_1(a)}{2\pi},\) hence \(n_0=n_1.\) Define the homotopy by \[\Ga_s(t)=z_0+[(1-s)r_0(t)+sr_1(t)]\exp(i(1-s)\th_0(t)+is\th_1(t)).\]
Clearly, \(\Ga_k=\ga_k\) for \(k=0,1.\) The only thing to check is that \(\Ga_s\) is a closed curve for each \(s\in[0,1],\) that is, to show \(\Ga_s(b)=\Ga_s(a).\) As \(r_k(a)=r_k(b),\) this is equivalent to \[\exp\bigl(i(1-s)(\th_1(b)-\th_1(a))+is(\th_0(b)-\th_0(a))\bigr)=1.\]
This holds since \(\th_1(b)-\th_1(a)=\th_0(b)-\th_0(a)=2\pi n_0\) and \[\exp\bigl(i(1-s)2\pi n_0+is2\pi n_0)=\exp(2\pi in_0\bigr)=1.\]
Theorem 10.1 (Residue theorem) Let \(U\subset\C\) be an open set, \(\ga\colon[a,b]\to U\) a null-homotopic piecewise C1 loop in \(U,\) and \(z_1,\ldots, z_n\in U\setminus\ga([a,b])\) points not on the loop \(\ga.\) If \(f\) is a holomorphic function on \(U\setminus\{z_1,\ldots, z_n\},\) then
\[\frac{1}{2\pi i}\int_\ga f(z)dz=\sum_{k=1}^n W_{z_k}(\ga)\Res_{z_k}(f).\]
Proof.
For each \(k=1,\ldots,n\) we find \(r_k>0\) and Laurent expansions \[f(z)=\sum_{n=-\iy}^{\iy}a_n^{(k)}(z-z_k)^n\qquad\text{for all $z\in D^\t_{r_k}(z_k).$}\]
Let \(P_k(z)=\sum_{n=-\iy}^{-1}a_n^{(k)}(z-z_k)^n\) be the corresponding principal part. Hence \(a_{-1}^{(k)}=\Res_{z_k}(f).\) By Proposition 9.1, \(P_k(z)\) is holomorphic on \(D^\t_{+\iy}(z_0).\) Also, \(f(z)-\sum_{k=1}^n P_k(z)\) defines a holomorphic function on \(U\) since its singularities at \(\{z_1,\ldots,z_n\}\) are removable. By Cauchy’s Theorem 7.1, \[\int_\ga f(z)dz=\sum_{k=1}^b\int_\ga P_k(z)dz.\]
By Proposition 10.2(b), the loop \(\ga\) is homotopic in \(\C\setminus\{z_k\}\) to \(\6 D_{r_k}(z_k)\ast\cdots\ast\6 D_{r_k}(z_k)\) (the same loop traversed \(W_{z_k}(\ga)\)-times). Combined with the fact that \(P_k\) is holomorphic on \(\C\setminus\{z_k\},\) this implies by Theorem 7.2 that \[\begin{align*} &\int_\ga P_k(z)dz=W_{z_k}(\ga)\int_{\6D_{r_k}(z_k)} P_k(z)dz \\ &=W_{z_k}(\ga)\int_{\6D_{r_k}(z_k)}\sum_{n=-\iy}^{-1}a_n^{(k)}(z-z_k)^ndz \end{align*}\] by uniform convergence, \[W_{z_k}(\ga)\int_{\6D_{r_k}(z_k)}\sum_{n=-\iy}^{-1}a_n^{(k)}(z-z_k)^ndz = W_{z_k}(\ga)\sum_{n=-\iy}^{-1}a_n^{(k)}\int_{\6D_{r_k}(z_k)}(z-z_k)^ndz\] using Equation 6.6 we have \[W_{z_k}(\ga)\sum_{n=-\iy}^{-1}a_n^{(k)}\int_{\6D_{r_k}(z_k)}(z-z_k)^ndz = W_{z_k}(\ga)a_{-1}^{(k)}2\pi i\] as required.
Determine the winding numbers \(W_{z_0}(\ga)\) in the following cases.
Use the residue theorem to compute \[\int_{\partial D_r(0)}\frac{(z+2)^2}{z(z-1)^2}dz\]
for all \(0<r\neq1.\)
Hint: Consider the cases \(r>1\) and \(r<1\) separately.
Apply the residue theorem to the contour \(\partial D_1(0)\) and a suitable holomorphic function \(f(z)\) to compute the integral\[\int_0^{2\pi}\frac{dx}{2+\cos(x)}.\]
Compute the integral \[\int_{0}^{+\iy}\frac{\cos(x)}{\sqrt{x}}dx\] by considering \(f(z)=\frac{e^{iz}}{\sqrt{z}}\) and the following contour \(\ga_{r,R}\) for \(r\to 0,\) \(R\to+\iy.\)
Hint: Recall that \(\int_0^{\iy}e^{-x^2}dx=\sqrt{\pi}/2.\)
Compute the integral \[\int_{-\iy}^{+\iy}\frac{dx}{1+x^6}\]
by applying the residue theorem to \(f(z)=\frac{1}{1+z^6}\) and the following upper semi-circular contour \(\ga_R\) for \(R\to+\iy.\)
